MTH101 Assignment No.1

Question 1: Discuss the continuity of the function f(x)=xtan(1/x) at x = 0 given that x≠0 and f(0)=0.

Now we check the continuity of f(x).

Here,

lim┬(x→0) f(x) = lim┬(x→0)⁡〖(xtan 1/x)^ 〗

lim┬(x→0) f(x) = lim┬(x→0)(x). lim┬(x→0)(tan 1/x)

lim┬(x→0) f(x) = 0 . lim┬(x→0)(tan 1/x)

lim┬(x→0) f(x) = 0

Hence! Limit of function exists at x = 0.

Since, given that f(0) is defined and f(0) = 0

So, we have

lim┬(x→0) f(x) = f(0) = 0

Therefore, by the definition of continuity we say that given function is continuous at

X = 0

Question 2: Check whether the given functions have at least one solution over the given intervals by using Intermediate value theorem.

Solution: (A)

Given that

f(x) = 2x3- 5x2 – 10x + 5

Here, (a = -1) and (b = 2)

Now,

f(-1) = 2(-1)3 – 5(-1)2 -10(-1) + 5

f(-1) = 8

and

f(2) = 2(2)3 – 5(2)2 -10(2) + 5

f(2) = -19

Since, f(-1) and f(2) are of opposite signs so by intermediate value theorem we conclude that there is at least one solution of f(x) in the interval [-1,2].

Solution (B)

Given that

h(x) = x4 + x -3

Here, (a = -2) and (b = 0)

Now,

h(-2) = (-2)4 + (-2) -3

h(-2) = 11

and

h(0) = (0)4 + (0) -3

h(0) = -3

Since, h(-2) and h(0) are of opposite signs so by intermediate value theorem we conclude that there is at least one solution of h(x) in the interval [-2,0].